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3 years ago

What did rutherfords gold foil experiment help him conclude

2 answers:

6 0

It allowed him to realize that the mass of an atom is concentrated at its center because the atoms mostly went through the foil but some were deflected. He also realized that an atom probably wasn't just empty space and scattered electron and it had a positive center.

Svetradugi [14.3K]3 years ago

3 0

Answer: the existence of the nucleus of an atom

Explanation:

Rutherford's Gold Foil Experiment proved the existence of a small massive center to atoms, which would later be known as the nucleus of an atom. Ernest Rutherford, Hans Geiger, and Ernest Marsden carried out their Gold Foil Experiment to observe the effect of alpha particles on the matter.

Most of the space inside the atom is empty because most of the α-particles passed through the gold foil without getting deflected. Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space.

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How many grams of kclo3 are needed to make 30.0 grams of kcl

lubasha [3.4K]

The mass of potassium chlorate, KClO₃ needed to produce 30 grams of potassium chloride, KCl is 49.33 grams

<h3>Balanced equation </h3>

2KClO₃ —> 2KCl + 3O₂

Molar mass of KClO₃ = 39 + 35.5 + (16×3) = 122.5 g/mol

Mass of KClO₃ from the balanced equation = 2 × 122.5 = 245 g

Molar mass of Kcl= 39 + 35.5 = 74.5 g/mol

Mass of KCl from the balanced equation = 2 × 74.5 = 149 g

SUMMARY

From the balanced equation above,

149 g of KCl were obtained from 245 g of KClO₃

<h3>How to determine the mass of KClO₃ needed </h3>

From the balanced equation above,

149 g of KCl were obtained from 245 g of KClO₃

Therefore,

30 g of KCl will be obtained from = (30 × 245) / 149 = 49.33 g of KClO₃

Thus, 49.33 g of KClO₃ are needed for the reaction

Learn more about stoichiometry:

brainly.com/question/14735801

7 0

2 years ago

Is KBr an acid or base

DiKsa [7]

Acid probably my guessing

5 0

4 years ago

calculate the number of moles of gas that occupy a 3.45L container at a pressure of 1.48 atm and a temperature of 45.6 Celsius

Otrada [13]

Answer:

There are 0,2 moles of gas that ocuppy the container.

Explanation:

We apply the formula of the ideal gases, we clear n (number of moles); we use the ideal gas constant R = 0.082 l atm / K mol. Firs we convert the unit of temperature in Celsius into Kelvin:

0°C= 273 K ------> 45,6 °C= 273 + 45, 6= 318, 6 K

PV= nRT ---> n= PV/RT

n= 1,48 atm x 3,45 L /0.082 l atm / K mol x 318,6 K

n= 0,195443479 mol

8 0

3 years ago

During studies of the reaction below,

Leni [432]

<u>Answer:</u> The percent yield of the nitrogen gas is 11.53 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For NO:</u>

Given mass of NO = 11.5 g

Molar mass of NO = 30 g/mol

Putting values in equation 1, we get:

$\text{Moles of NO}=\frac{11.5g}{30g/mol}=0.383mol$

<u>For $N_2O_4$ :</u>

Given mass of $N_2O_4$ = 102.1 g

Molar mass of $N_2O_4$ = 92 g/mol

Putting values in equation 1, we get:

$\text{Moles of }N_2O_4=\frac{102.1g}{92g/mol}=1.11mol$

For the given chemical reactions:

$2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)$ ......(2)

$N_2H_4(l)+2N_2O_4(l)\rightarrow 6NO(g)+2H_2O(g)$ .......(3)

<u>Calculating the experimental yield of nitrogen gas:</u>

By Stoichiometry of the reaction 3:

6 moles of NO is produced from 2 moles of $N_2O_4$

So, 0.383 moles of NO will be produced from = $\frac{2}{6}\times 0.383=0.128mol$ of $N_2O_4$

By Stoichiometry of the reaction 2:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 0.128 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 0.128=0.384mol$ of nitrogen gas

Now, calculating the experimental yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 0.384 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$0.384mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.384mol\times 28g/mol)=10.75g$

<u>Calculating the theoretical yield of nitrogen gas:</u>

By Stoichiometry of the reaction 2:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 1.11 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 1.11=3.33mol$ of nitrogen gas

Now, calculating the theoretical yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 3.33 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$3.33mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(3.33mol\times 28g/mol)=93.24g$

To calculate the percentage yield of nitrogen gas, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of nitrogen gas = 10.75 g

Theoretical yield of nitrogen gas = 93.24 g

Putting values in above equation, we get:

$\%\text{ yield of nitrogen gas}=\frac{10.75g}{93.24g}\times 100\\\\\% \text{yield of nitrogen gas}=11.53\%$

Hence, the percent yield of the nitrogen gas is 11.53 %.

7 0

3 years ago

40. Write a balanced equation and a net ionic equation for combining AgNO3(aq) and Na2CO3(aq)

ki77a [65]

2AgNO3 (aq) + Na2CO3 (aq) --> Ag2CO3 (s) + 2NaNO3 (aq)

2Ag(+) (aq) + CO3(2-) (aq) --> Ag2CO3 (s)
hope it helps

3 0

4 years ago